# How do we know that wave f. is the eigenfunction of an operator H?

1. May 10, 2013

### 71GA

I am kind of new to this eigenvalue, eigenfunction and operator things, but i have come across this quote many times:
First i need some explaination on how do we know this? All i know about operator $\hat{H}$ so far is this equation where $\langle W \rangle$ is an energy expected value:

\begin{align}
\langle W \rangle &= \int \limits_{-\infty}^{\infty} \overline{\Psi}\, \left(- \frac{\hbar^2}{2m} \frac{d^2}{d \, x^2} + W_p\right) \Psi \, d x
\end{align}

From which it follows that $\hat{H} = - \frac{\hbar^2}{2m} \frac{d^2}{d \, x^2} + W_p$.

I know how to derive relation $\hat{H}\hat{a} = (W - \hbar \omega)\hat{a} \psi$ for which they state that:

I also know how to derive relation $\hat{H}\hat{a}^\dagger = (W + \hbar \omega)\hat{a}^\dagger \psi$ for which they state that:

How do we know this?

2. May 10, 2013

### HallsofIvy

Staff Emeritus
That's pretty much the definition of "eigenfunction" and "eigenvalue", isn't it?

3. May 10, 2013

### 71GA

What exactly here is the definition? Which equation?

4. May 10, 2013

### Fightfish

Any equation of the form
$\hat{H} \psi = E \psi$
is an eigen-equation

The functions $\psi_n$ which satisfy this equation are called the eigenfunctions of the operator $\hat{H}$, with corresponding eigenvalues $E_n$.

Put simply, when an operator $\hat{H}$ acts upon one of its eigenfunctions, it returns the same eigenfunction multiplied by a number, which we call the eigenvalue.

5. May 11, 2013

### 71GA

1st:
In this equation $\hat{H}\hat{a} = (W - \hbar \omega)\hat{a} \psi$ is it possible that $\hat{a} \psi$ is an eigenfunction... i mean $\hat{a}$ is an operator...

2nd:
And is it possible that my quotings were wrong i will write the wrong one and the one that i think is better:

WRONG:
$\hat{a}\psi$ is an eigenfunction of operator $\hat{H}$ with eigenvalue $(W−\hbar \omega)$.

RIGHT:
$\hat{a}\psi$ is an eigenfunction of operators $\hat{H}\hat{a}$ with eigenvalue $(W−\hbar \omega)$.

6. May 11, 2013

### Staff: Mentor

What's right is wrong and what's wrong is right
An operator maps one function into another, so if $\hat{a}$ is an operator and $\psi$ is a function, then $\hat{a}\psi$ is itself a function.

And I think that you dropped a $\psi$ from the equation you started with. I would have expected $\hat{H}\hat{a}\psi = (W - \hbar \omega)\hat{a} \psi$, and written that way it is clear that the function $\hat{a}\psi$ is an eigenfunction of $\hat{H}$ with eigenvalue $(W−\hbar \omega)$... Or at least adding a few parentheses will make it clear:

$$\hat{H}(\hat{a}\psi) = (W - \hbar \omega)(\hat{a} \psi)$$

7. May 12, 2013

### 71GA

Ok so the eigenfunction can be a combination of an operator $\hat{a}$ and a function $\psi$?

8. May 12, 2013

### Fightfish

Why not? An operator acting on a function gives you another function.

9. May 12, 2013

Thanks.